A man mixed 10 litre of water at 50 degrees temperature with 20 litre of water such that the mixture is at 30 degree temperature. What is the temperature of the 20 litre water?
a) 10 degree
b) 20 degree
c) 25 degree
d) 30 degree
Share
A man mixed 10 litre of water at 50 degrees temperature with 20 litre of water such that the mixture is at 30 degree temperature. What is the temperature of the 20 litre water?
a) 10 degree
b) 20 degree
c) 25 degree
d) 30 degree
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
The temperature of the 20 litre water = 20 degrees
Given:
A man mixed 10 litre of water at 50 degrees temperature with 20 litre of water
The temperature of resultant mixture is at 30 degree
To find:
What is the temperature of the 20 litre water
Solution:
Let's assume that V₁ and V₂ are the volume of initial mixtures, T₁ and T₂ are be the temperatures
And Tₓ be the temperature of the resultant mixture
According to given data,
V₁ = 10 liters, V₂ = 20 liters, T₁ = 50 degrees, and Tₓ = 30 degrees
Now we need find T₂
Here, the heat gained by the first volume of water = V₁ × C × (Tₓ -T₁)
the heat lost by the 2nd volume of water = V₂ × C × (T₂ - Tₓ)
By the law of conservation of energy,
heat lost = heat gained
⇒ V₁ × C × (Tₓ -T₁) = V₂ × C × (T₂ - Tₓ)
⇒ V₁ × (Tₓ -T₁) = V₂ × (T₂ - Tₓ)
⇒ V₁Tₓ - V₁T₁ = V₂T₂ - V₂Tₓ
⇒ V₂T₂ = V₂Tₓ + V₁Tₓ - V₁T₁
⇒ T₂ = (V₂Tₓ + V₁Tₓ - V₁T₁)V₂
From above data
T₂ = (V₂Tₓ + V₁Tₓ - V₁T₁) / V₂
⇒ T₂ = (20(30) + (10)(30) - (10)(50) / 20
⇒ T₂ = 600 + 300 - 500 / 20
⇒ T₂ = 400 / 20 = 20
The temperature of the 20 litre water = 20 degrees
#SPJ1